Q. The length of a cylinder is measured with a meter rod having least count $0.1cm$ . Its diameter is measured with Vernier calipers having least count $0.01cm$ . Given that length is $5.0$ $cm$ and radius is $2.0cm$ . What will be the percentage error in the calculated value of the volume ?
NTA AbhyasNTA Abhyas 2020
Solution:
Assuming $r$ as the radius of cylinder , $d$ as the diameter and $l$ as its length
since $d=2r\Rightarrow \Delta r=\Delta d$
since least count is also the maximum permissible error
$\Rightarrow \frac{\Delta r}{r}=\frac{0 . 01}{2 . 0};\frac{\Delta l}{l}=\frac{0 . 1}{5 . 0}$
Volume of cylinder $V=\pi r^{2}l$
Percentage error in volume
$\Rightarrow \frac{\Delta V}{V}\times 100=\frac{2 \Delta r}{r}\times 100+\frac{\Delta l}{l}\times 100$
$=\left(2 \times \frac{0 .01}{2 .0} \times 100 + \frac{0 .1}{5 .0} \times 100\right)=\left(1 + 2\right)\%=3\%$
since $d=2r\Rightarrow \Delta r=\Delta d$
since least count is also the maximum permissible error
$\Rightarrow \frac{\Delta r}{r}=\frac{0 . 01}{2 . 0};\frac{\Delta l}{l}=\frac{0 . 1}{5 . 0}$
Volume of cylinder $V=\pi r^{2}l$
Percentage error in volume
$\Rightarrow \frac{\Delta V}{V}\times 100=\frac{2 \Delta r}{r}\times 100+\frac{\Delta l}{l}\times 100$
$=\left(2 \times \frac{0 .01}{2 .0} \times 100 + \frac{0 .1}{5 .0} \times 100\right)=\left(1 + 2\right)\%=3\%$